Problem: Simplify; express your answer in exponential form. Assume $t\neq 0, q\neq 0$. $\dfrac{{(t^{2})^{-3}}}{{(t^{-5}q^{3})^{4}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${t^{2}}$ to the exponent ${-3}$ . Now ${2 \times -3 = -6}$ , so ${(t^{2})^{-3} = t^{-6}}$ In the denominator, we can use the distributive property of exponents. ${(t^{-5}q^{3})^{4} = (t^{-5})^{4}(q^{3})^{4}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(t^{2})^{-3}}}{{(t^{-5}q^{3})^{4}}} = \dfrac{{t^{-6}}}{{t^{-20}q^{12}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{-6}}}{{t^{-20}q^{12}}} = \dfrac{{t^{-6}}}{{t^{-20}}} \cdot \dfrac{{1}}{{q^{12}}} = t^{{-6} - {(-20)}} \cdot q^{- {12}} = t^{14}q^{-12}$.